(MM)C5H12=5x12+12x1=72g/mol (MM)O2=2x16=32g/mol Acoording to equation: C5H12 + 8O2 => 5CO2 + 6H2O
The proportion between C5H12 and O2 is 1:8 in mol
So: 72/16=46/x finale x=10,2g of O2
To combust 46.0 grams of pentane, approximately 163.33 grams of oxygen are required. This is calculated using the chemical equation for combustion, molar mass conversions, and mole ratios. The steps include determining the molar mass, converting mass to moles, and using the mole ratio to find the required mass of oxygen.
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